Remarks on R2

R2 depends on the variance on the variance of the predictors

Quoting from Shalizi¹ Assuming a true linear model
\[ Y = aX + \epsilon\]
and assuming we know \(a\) exactly.
The variance of Y will be \(a^2\mathbb{V}[X] + \mathbb{V}[\epsilon]\).
So \(R^2 = \frac{a^2\mathbb{V}[X]}{a^2\mathbb{V}[X] + \mathbb{V}[\epsilon]}\)
This goes to 0 as \(\mathbb{V}[X] \rightarrow 0\) and it goes to 1 as \(\mathbb{V}[X] \rightarrow \infty\). “It thus has little to do with the quality of the fit, and a lot to do with how spread out the predictor variable is. Notice also how easy it is to get a high \(R^2\) even when the true model is not linear!”

Below a quick comparison between two linear relationships, one with much higher variance than the other in the predictor.
Added a different constant for better display in plot.

library(tidyverse)

x1 = rnorm(1000, mean=0, sd=1)
x2 = rnorm(1000, mean=0, sd=10)
error = rnorm(1000, mean=0, sd=0.5)

y1 = x1 + error
y2 = 10 + x2 +  error

df = data.frame(x1,x2,y1,y2)

model1 = lm("y1 ~ x1")
model2 =  lm("y2 ~ x2")

R2 of lower variance predictor: 0.8

R2 of higher variance predictor: 1